Question

Suppose a pendulum of length L meters makes an angle of θ radians with the vertical, as in the figure. It can be shown that as a function of time,θ satisfies the differential equation d²θ/dt² + g/L sin θ = 0, where g = 9.8 m/s² is the acceleration due to gravity. For theta near zero we can use the linear approximation sinθ is approximately equal to theta to get a linear differential equation d²θ/dt² + g/Lθ = 0. Use the linear differential equation to answer the following questions. (a) Determine the equation of motion for a pendulum of length 1.5 meters having initial angle 0.4 radians and initial angular velocity dθ/dt = 0.2 radians per second. θ(t) = radians (b) What is the period of the pendulum? That is, what is the time for one swing back and forth? Period = seconds.

Answer 1

Using the linear approximation for small oscillations, we solve for θ(t) with initial conditions θ(0) = 0.4 radians and dθ/dt (0) = 0.2 rad/s. We determine the angular frequency to find the equation of motion, and use that to calculate the period.

To find the equation of motion for the pendulum, we will be using the linear differential equation d²θ/dt² + (g/L)θ = 0, where g is the acceleration due to gravity (9.8 m/s²) and L is the length of the pendulum (1.5 meters).

First, we recognize this equation resembles the harmonic oscillator equation whose solution has the form θ(t) = Acos(ωt - δ), where ω is the angular frequency and A and δ are determined from initial conditions. Using the initial conditions θ(0) = 0.4 radians and dθ/dt (0) = 0.2 radians per second, we find A and δ. The angular frequency ω can be found from ω² = g/L, leading to ω = √(g/L). Substituting g = 9.8 m/s² and L = 1.5 meters, the angular frequency is ω = √(9.8/1.5).

The period T of the pendulum is found using the formula T = 2π/ω. Plugging in the calculated ω gives us the period for one complete oscillation.